Probability of two independent events not occuring Question - The probability that at least one of the two
independent events occurs is 0.5. The probability that the first event occurs but not the second is
3/25. Also, the probability that the second event occurs but not the first is 8/25. Find the
probability that none of the two events occurs.
Through this formula P(A'∩ B') = P(A'). P(B'), I am getting 0.5084, But if I complement P(AUB) which is given 0.5, It is obvious that is also 0.5. So, why am I getting different answers for two same things essentially?
Edit: If anyone is curious how I got 0.5084
P(A-B) = P(A) - P(A∩ B) -> 1
P(B-A) = P(B) - P(A∩ B) -> 2
P(AUB) = P(A)+P(B)-P(A∩ B) 
P(B) = 0.5 - 3/25 = 0.18 [from 1]
and similarly P(A) = 0.5-8/25 = 0.38
P(A'∩ B') = P(A'). P(B')
= 0.82*0.62 = 0.5048
Thank you so much to all of you and here I was banging my head in the wall thinking what am I doing wrong, I wasn't expecting the quick replies so thanks again and I think I would go with the 0.5.
 A: You are correct that the problem statement gives contradictory information.  The probabilities given do not add up correctly.
You correctly found that $\Pr(A'\cap B')=1-\Pr(A\cup B)=1-0.5=0.5$ by DeMorgan's Laws.
You likely found that $\Pr(B') = \Pr(A'\cap B')+\Pr(A\cap B') = 0.5+\frac{3}{25}$ by the Law of Total Probability.
You similarly would have found that $\Pr(A')=\Pr(A'\cap B')+\Pr(A'\cap B) = 0.5+\frac{8}{25}$
You correctly found that if the events $A$ and $B$ were in fact independent then $A'$ and $B'$ would similarly be independent and that $\Pr(A'\cap B') = \Pr(A')\times\Pr(B')$ by definition of independence which would have required then that $(0.5+\frac{3}{25})\times(0.5+\frac{8}{25})=\Pr(A'\cap B')$ however this equaled $0.5084$ which does not match up with the $0.5$ we expected.
One of two things has likely gone wrong: The problem gave us incorrect numbers, perhaps due to rounding where not appropriate, for the values given in the problem... or they incorrectly told us that the events were independent when in fact they were not.
A: We are given $P(A \cup B) = 0.5, P(A \cap B') = 0.12$ and $P(A' \cap B) = 0.32$.
$P(A) = P(A \cup B) - P(A' \cap B) = 0.5 - 0.12 = 0.38$
$P(B) = P(A \cup B) - P(A \cap B') = 0.5 - 0.32 = 0.18$
Going by the formula, $P(A') = 0.62, P(B') = 0.82$ and $P(A' \cap B') = 0.62 \cdot 0.82 = 0.5084$. So both of your answers are correct. The question seems to be poorly constructed.
A: Ther might be a mistake in the formulation: could it be that instead of "the probability that at least one of the two independent events occurs is 0.5" it should be "the probability that both of the two independent events occurs is 0.5"? If so, then the following approach solves your problem.
Fot two independent events $A$ and $B$ that might happen, there are only 4 possible outcomes of the experiment:
$$
\begin{aligned}
AB &= \text{"both }A\text{ and }B\text{ occur"} \\
\overline{A}B &= \text{"}A\text{ doesn't occur }\text{ and }B\text{ occurs"} \\
A\overline{B} &= \text{"}A\text{ occurs }\text{ and }B\text{ doesn't occur"} \\
\overline{A}\overline{B} &= \text{"neither }A\text{ no }B\text{ occur"} \\
\end{aligned}
$$
Then, we have that
$$
\begin{aligned}
P(AB + \overline{A}B + A\overline{B}+\overline{A}\overline{B}) &= P(AB) + P(\overline{A}B) + P(A\overline{B})+P(\overline{A}\overline{B}) = 1 \Rightarrow \\
\Rightarrow P(\overline{A}\overline{B}) &= 1 - P(AB) - P(\overline{A}B) - P(A\overline{B}) = \\
&= 1 - 0.5 - \frac{8}{25} - \frac{3}{25} = \frac{3}{50}. 
\end{aligned}
$$
